Prove:
G_{1} en G_{2} are isomorphic if and only if the multiplication tables are the same
How do you prove something like this?
well, for starters, it's not quite true. you have to add the caveat: "the same (up to a reordering of the rows and columns)".
secondly, such a table exists only if both groups are FINITE (one can "imagine" infinite tables, but how does one write them down? for example, what if G_{1} is the additive group of the real numbers? (there is actually a "trick" for this, involving rational sums, but that's besides the point)).
but, assuming G_{1}, G_{2} are finite, if the multiplication tables are "the same" (except for a "re-labelling" of the elements), then let the elements of G_{1} be: {e_{1},g_{1},...,g_{n}} and let the elements of G_{2} be {e_{2},h_{1},...,h_{n}}, where the elements are listed in the order they occur at the top of the columns of their respective tables.
define the mapping: φ:G_{1}→G_{2} by:
φ(e_{1}) = e_{2}
φ(g_{j}) = h_{j}, for j = 1,2,...,n.
then φ(g_{j}g_{k}) = h_{j}h_{k} = φ(g_{j})φ(g_{k}) (since the tables are the same).
thus φ is a homomorphism, and it is clearly bijective. thus G_{1} and G_{2} are isomorphic (since φ is an isomorphism).
(i've omitted the products involving the identity elements, which you should also verify. this isn't hard (the "top row of the first table" maps to the "top row of the second table" and similarly for the "left-most column")).
on the other hand, suppose G_{1} and G_{2} are isomorphic, so that we HAVE an isomorphism φ already.
then replace every element x in the table for G_{1}, with φ(x), and voila! we have a table for G_{2}.
so why do we deal with isomorphisms, instead of just "comparing tables"? because for a group G of order n, the multiplication table for G has n^{2} entries, which is often quite large compared to n (a group of order 12, has 144 entries to check). on the other hand, if we can establish that a bijective map is a homomorphism (and thus an isomorphism), we often have a one or two-line proof that two groups are isomorphic, which is a LOT easier than comparing two tables of large size to each other "one entry at a time." that is, isomorphisms let us "compare tables" much more efficiently.
also: as a practical matter, it may not be obvious that the rows (columns) of two tables can be re-arranged so that they're "the same". again, as the order of G (say n) gets very large, there are n! possible ways to re-arrange the rows (or, equivalently, the columns), and n! grows even faster than n^{2}. while for groups of small order (like 3 or 4), it may be feasible to check "all orderings of the elements of G", for a group of order 12 there are 479,001,600 (that's more than 479 MILLION) ways we might "re-arrange the rows (or columns)". it's just not humanly possible for a person to check all those (i suppose a computer could). so while a "multiplication table" might seem like a good way to describe a group operation conceptually, in practice it's not very convenient (except with "baby groups").
this is just my personal feeling on the matter, but i feel that the emphasis on "tables" is a poor introduction to groups. mathematicians do not, in general, specify a group by exhibiting a cayley table. they would be of rather limited value if that was the case. if i may make an analogy: one does not get a "feel" for how the sine function behaves, by looking at a table of values. rather, one notices:
a) sine is periodic, it repeats every 2π (2*pi)
b) sine varies from -1 to 1 in value
c) sin(0) = 0, sin(π/2) = 1, sin(π) = 0, sin(3π/2) = -1
d) sin(-x) = -sin(x)
etc.
these describe the "qualitative" behavior of the function sine. with groups, it's the same way: the "qualitative" behavior of groups is what makes them interesting and useful (is a group cyclic? abelian? finite? among other things), which the tables don't often convey very well. it's far easier to see the symmetry in a and b of the equation:
ab = ba (if we "swap" a and b, we get the equation ba = ab, which says the same thing, since equality ITSELF is symmetric),
than it is to notice the corresponding symmetry in a table (above and below the diagonal), especially if the table itself is very large. in a similar vein, one does not usually prove a relation on a set S is an equivalence relation by "drawing it on SxS", but rather from the PROPERTIES an equivalence relation MUST have. and this is what isomorphisms essentially ARE: "group-property-preserving" maps.
Thanks man! Really helpful informative post !
I understand that for those large finite groups, multiplication tables are quite complicated
But, for those baby groups it can be helpful I guess ?
For example, we had another exercise were you had to prove that groups of order 3 are isomorphic.
I proved this by showing that the multiplication tables have to be the same.
Seems quite hard to me to prove this without using multiplication tables..
actually, it's not. let G = {e,a,b} and H = {1,x,y} be two groups of order 3. i claim that G = <a> = <b>, and that H = <x> = <y>, and that any of the following maps are isomorphisms:
e-->1
a-->x
b-->y
e-->1
a-->y
b-->x
consider a^{2}. by closure a^{2} must be some element of G. if a^{2} = e, then a = a^{-1}. this means b^{-1} = b, as well (since e is its own inverse, and since a can't be b's inverse, since it is already a's inverse). also by closure, ab must be in G. now:
ab = e means b = a^{-1}, but a = a^{-1}, so ab is not e.
ab = a means b = e
ab = b means a = e.
so ab is a fourth element of G, contradicting the fact that G only has 3 elements. so it must not be the case that a^{2} = e. since a^{2} = a means a = e (which isn't true, since a and e are different elements of G), it must be the case that a^{2} = b. a similar analysis show that a^{-1} must also be b:
if a^{-1} = e, then a = (a^{-1})^{-1} = e^{-1} = e, impossible.
if a^{-1} = a, then e = aa^{-1} = a^{2}, which we have already ruled out.
hence a^{3} = a(a^{2}) = ab = aa^{-1} = e. that is G = {e,a,a^{2}} = <a>.
since a^{-1} = b, we have that b^{-1} = (a^{-1})^{-1} = a, and
b^{2} = (a^{-1})^{2} = (a^{2})^{-1} = b^{-1} = a (from above).
thus b^{3} = (b^{2})b = ab = aa^{-1} = e, as well, so G = {e,b,b^{2}} = <b>.
the exact same argument applied to H shows that H = <x> = <y>, and that xy = 1. thus the first mapping i gave above is:
e= a^{3}-->x^{3} = 1
a--> x
b = a^{2}--> x^{2} = y
while the second is:
e = b^{3}-->x^{3} = y^{3} = 1
a = b^{2}-->x^{2} = y
b --> x
so that we have either φ(a^{k}) = x^{k}, or φ(b^{k}) = x^{k} (so that φ is a homomorphism by the rules of exponents).
*******
a "real group-theorist" wouldn't even go to all this trouble. she would simply remark that 3 is prime, so both G and H are cyclic of order 3, and both isomorphic to the integers mod 3 under addition (i didn't do this, because it is not clear if you have covered lagrange's theorem yet, which has PROFOUND implications for what a group can, and cannot be).